数学Paper 1讨论专区

whyyie

walrein_lim88

回复 1541# whyyie

whyyie

回复 1542# walrein_lim88

第八行不应该cancel掉n, 这样就少掉了一个solution.
可以解释part (b) 吗? 不是很明白

Allmaths

回复  walrein_lim88

第八行不应该cancel掉n, 这样就少掉了一个solution.
可以解释part (b) 吗? 不是很 ...
whyyie 发表于 16-8-2010 10:09 PM

   以我看法，n 是不可能等于0..所以删除掉也不会影响final answer的..最终final answer拿到0也是要reject掉的..

whyyie

回复 1544# Allmaths

When n =0,
a=0; d=0
=> arithmetic progression


错了?

Allmaths

回复  Allmaths

When n =0,
a=0; d=0
=> arithmetic progression


错了?
whyyie 发表于 16-8-2010 10:47 PM

没见过a=0和d=0的AP..虽然书的答案有0...

还有你part (b) 的题目打错了吧..

whyyie

回复 1546# Allmaths

还真的打错了, 没去注意

Allmaths

回复  Allmaths

还真的打错了, 没去注意
whyyie 发表于 16-8-2010 11:17 PM

walrein_lim88

哈哈。。看来这里真的是靠allmaths了。。加油！！！

Allmaths

哈哈。。看来这里真的是靠allmaths了。。加油！！！
walrein_lim88 发表于 17-8-2010 07:57 AM

不敢..这题目是STPM PAST YEAR...所以才会做..

芭樂

回复 1539# peaceboy


   
这个是答案
anyway 谢谢

白羊座aries

1.find the value of t such that 2y=x+t is a tangent to the ellipse x^2+16y^2=16


2. A straight line 2y=x+1 intersects the hyperbola 4x^2-9y^2 = 36 at 2 points P and Q . Find the coordinates of the midpoint of PQ .

3.Find the equation of the tangent to the hyperbola x^2-2y^2=1 which is paralllel to the line 4y=3x

芭樂

回复 1552# 白羊座aries


   
应该是这样...
不过感觉好像有点错

peaceboy

1.find the value of t such that 2y=x+t is a tangent to the ellipse x^2+16y^2=16


2. A straight l ...
白羊座aries 发表于 17-8-2010 08:31 PM

    1.find the value of t such that 2y=x+t is a tangent to the ellipse x^2+16y^2=16

x^2+16y^2=16
2x + 32y(dy/dx) =0
m=dy/dx = -2x/32y

y=mx+c
y=x/2+t/2

when the gradient of tangent and the gradient point of ellipse are the same ..

-2x/32y = 1/2
-4x=32y
8y+x=0
x=-8y

sub x=-8y into x^2+16y^2=16 to get the point of the tangent
64y^2 + 16y^2 = 16
80y^2=16
y= +- (1/5)^(1/2)
x= - + 8(1/5)^(1/2)


from the ques
2y=x+t , sub back the point of tangent into the equation
y= + (1/5)^(1/2) ,x= -  8(1/5)^(1/2)
2(1/5)^(1/2) = -8(1/5)^(1/2) +t
t=10(1/5)^(1/2)

y=- (1/5)^(1/2) , x=+ 8(1/5)^(1/2)
-2 (1/5)^(1/2) =+ 8(1/5)^(1/2) + t
t=-10(1/5)^(1/2)



3.Find the equation of the tangent to the hyperbola x^2-2y^2=1 which is paralllel to the line 4y=3x

y=3x/4
m=3/4

x^2-2y^2=1
2x - 4y(dy/dx) =0
dy/dx = x/2y

the same gradient of tangent and hyperbola ..
2x/4y=3/4
8x=12y
2x=3y
x=3y/2


sub x=3y/2 into x^2-2y^2=1 to get the point
9y^2/4 -2y^2 =1
9y^2 -8y^2 =4
y^2=4
y= +- 2
x=3y/2, x=-+3


we know that if the line is parallel , the gradient is the same , so we take t as a unknow that can change the position of line... and we sub back the pnt we get ...
y=3x/4 + t
x=-3 , y= 2
t=2 + 9/4
= 17/4

x=3 , y=-2
-2 = 9/4 +t
t= - 17/4

eqn , 1 . y=3x/4 + 17/4
            4y=3x+17
        2.y=3x/4 - 17/4
           4y=3x-17

白羊座aries

1.Given 2 points A ( -5,-8 ) and C ( 3,4) .A straight line which passes through the point A touches a circle with C as the centre at P. find the equation of AP and the equation of the circle

2.Find the equation of the circle with radius 5 units,which touch the y-axis and pass through the point (3,1)

peaceboy

1.Given 2 points A ( -5,-8 ) and C ( 3,4) .A straight line which passes through the point A touches  ...
白羊座aries 发表于 18-8-2010 12:54 AM

    第一题有答案么？不是很清楚问题说什么

2.2.Find the equation of the circle with radius 5 units,which touch the y-axis and pass through the point (3,1)

let the center = (a,b)

(x-a)^2 + (y-b)^2 = 25
x=3 , y=1
(3-a)^2 + (1-b)^2 = 25
9-6a+a^2 + 1-2b+b^2 = 25
a^2 - 6a + b^2 -2b = 15 ---- 1

the point touch the y-axis ,(0,b)
(0-a)^2 + (b-b)^2 = 25
a^2 = 25
a= +-5

from 1 , when a = 5 ,
a^2 - 6a + b^2 -2b = 15
25-30+b^2-2b = 15
b^2-2b = 20
b^2-2b-20=0
b= (2+-(4+80)^(1/2)) / 2
  =[ 2+- 2(21)^(1/2) ] / 2
  = 1 +- (21)^(1/2)

when a = -5
(-5)^2 - 6(-5)+ b^2 -2b = 15
25+30+ b^2 -2b = 15
b^2-2b+40=0
b^2 - 4ac = 4-4(1)(40)
              = -76  (<0) , have no real roots

so the equation is (x-5)^2 + (y-[1+(21)^(1/2)])^2 = 25
and (x-5)^2 + (y - [1-(21)^(1/2)])^2 = 25

芭樂

If y^3=6xy-x^3-1, prove that dy/dx=(2y-x^2)/(y^2-2x) and that the maximum value of y occurs when x^3=8+√114 and the minimum value when x^3=8-√114 .

peaceboy

If y^3=6xy-x^3-1, prove that dy/dx=(2y-x^2)/(y^2-2x) and that the maximum value of y occurs when x^3 ...
芭樂 发表于 18-8-2010 11:41 PM

    y^3=6xy-x^3-1
3y^2(dy/dx) = 6x(dy/dx) + 6y -3x^2
3y^2(dy/dx)- 6x(dy/dx)=6y -3x^2
dy/dx =  (6y -3x^2)/(3y^2-6)
         =3(2y-x^2)/3(y^2-2x)
         =(2y-x^2)/(y^2-2x)

dy/dx=0
(2y-x^2)/(y^2-2x)=0
(2y-x^2)=0
x^2 = 2y
y=(x^2)/2

sub back to the equation .,  y^3=6xy-x^3-1
  ((x^2)/2)^3=6x((x^2)/2)-x^3-1
x^6/8 = 6x^3/2 - x^3 - 1
x^6 - 16x^3 +8 =0

x^3 = [16 +-(16^2 - 4(1)(8)) ]/2
      = 8 +- (224)^(1/2) / 2
      = 8 +- (56)^1/2 ...

to find out maximun or min , use the table method ,

答案不太一样 = =你帮我检查看

强

http://www.scribd.com/doc/36071303/Chapter-Past-Year-Question

帮忙宣传一下。。。这是本人自己打字的。。2002-2009的往年试题，那些没有但是想要的可以下载来练习。。。

芭樂

1 A right circular cone of height h is inscribed in a sphere of radius R. Show that the volume V of the cone is given by
V = (pie/3)(2Rh^2-h^3)

2 A right circular cone of base radius r and height h has a total surface area S and volume V.Show that 9V^2 = r^2(S^2-2pier^2S). Hence or otherwise, show that for a fixed surface area S, the maximum volume of the cone occurs when its   semi-vertical angle A is given that by tan A = 8^-.5

3 A right pyramid has a square base of side x m and a total surface area (base and four sides) 72m^2. Show that the volume, Vcm^3, is given by  v^2 = 144x^2 - 4x^4. If x varies, find the value of x for which V is a maximum and obtain the maximum values of V.

谢谢。帮我prove就可以了